1 Introduction
Throughout the article, all operators are bounded linear operators acting on complex Hilbert spaces. We denote by
$\mathbb {D} , \mathbb {T} , r\mathbb {D}, r\mathbb {T}$
the unit disk, the unit circle, the disk with radius r and the circle with radius r respectively with center at the origin in the complex plane
$\mathbb {C}$
. For a Hilbert space
${\mathcal {H}}$
, we mean by
$\mathcal B({\mathcal {H}})$
the algebra of operators acting on
${\mathcal {H}}$
. A contraction is an operator whose norm is not greater than
$1$
. For
$0<r<1$
, let us consider the following annuli:
$$ \begin{align} A_r & =\{ z \in \mathbb{C} \,:\, r<|z|<1 \} \ , \notag \\ \mathbb A_r &= \{ z\in \mathbb{C} \, : \, r< |z| < {r}^{-1} \}. \end{align} $$
A Hilbert space operator T is said to be an
$A_r$
-contraction if
$\overline {A}_r$
is a spectral set for T, which is to say that the spectrum
$\sigma (T) \subseteq \overline {A}_r$
and for every rational function f with poles off
$\overline {A}_r$
von Neumann’s inequality holds, i.e.,
$ \displaystyle \|f(T)\| \leq \sup \, |f(z)|, $
where the supremum is taken over
$\overline {A}_r$
. Here
$f(T)=p(T)q(T)^{-1}$
, whence
$f=p/q$
with
$p,q \in \mathbb {C}[z]$
and q having no zeros in
$\overline {A}_r$
. An
$A_r$
-unitary is a normal operator having its spectrum in the boundary
$\mathbb {T} \cup r \mathbb {T}$
of the annulus
$A_r$
. In the seminal paper [Reference Agler1], Agler proved that every
$A_r$
-contraction dilates to an
$A_r$
-unitary which announces the success of rational dilation on an annulus. Hence, for any
$A_r$
-contraction T acting on a Hilbert space
${\mathcal {H}}$
, there is a Hilbert space
$\mathcal K \supseteq {\mathcal {H}}$
and an
$A_r$
-unitary N such that
$f(T)=P_{{\mathcal {H}}}f(N)|_{{\mathcal {H}}}$
for every rational function f with poles off
$\overline {A}_r$
. This path-breaking work due to Agler motivates numerous mathematicians to study further the functions and operators associated with an annulus which leads to exciting works like [Reference Agler, Harland and Raphael2, Reference Dritschel, Jury and McCullough6–Reference McCullough and Sultanic11]. Also, see the references therein. Recently, Bello and Yakubovich [Reference Bello and Yakubovich4] introduced two important classes of operators associated with an annulus, namely the
$C_{\alpha }$
and
$C_{1,r}$
classes which were defined in the following way:
$$ \begin{align*} C_{\alpha} &= \{T : T \text{ is an invertible operator and } \, \alpha(T) {\kern-1pt}=-{T^*}^2T^2{\kern-1pt}+{\kern-1pt}(1{\kern-1pt}+{\kern-1pt}r^2)T^*T-r^2I \geq 0 \}, \\C_{1,r} &= \{T : T \text{ is an invertible operator and } \, \|T\|, \, \|rT^{-1}\| \leq 1 \}. \end{align*} $$
We mention two important facts about these classes: first, the operators in
$C_{\alpha }, \, C_{1,r}$
classes have their spectrums in
$\overline {A}_r$
and second, if
$\mathcal A_r$
denotes the set of all
$A_r$
-contractions, then
$\mathcal A_r, \, C_{\alpha }, \, C_{1,r}$
form a strictly increasing chain as was proved in [Reference Bello and Yakubovich4].
Theorem 1.1 ([Reference Bello and Yakubovich4], Theorem 1.1)
$\mathcal A_r \subsetneq C_\alpha \subsetneq C_{1, r}$
.
Also, an explicit model was constructed for an operator in the
$C_{\alpha }$
class by Bello and Yakubovich, see [Reference Bello and Yakubovich4, Theorem 1.2]. Interestingly, McCullough and Pascoe [Reference McCullough and Pascoe12] considered the annulus
$\mathbb A_r$
as in (1.1) and introduced the quantum annulus
$\mathbb Q \mathbb A_r$
which consists of invertible operators T such that both
$rT$
and
$rT^{-1}$
are contractions, i.e.,
$$\begin{align*}\mathbb Q \mathbb A_r = \{T\,:\, T \text{ is an invertible operator and } \, \|T\|, \, \|T^{-1}\| \leq r^{-1} \}. \end{align*}$$
Evidently,
$C_{1,r} \subsetneq \mathbb Q \mathbb A_r$
which stretches the increasing chain of Theorem 1.1 one more step. In [Reference McCullough and Pascoe12], McCullough and Pascoe found the following model theorem for an operator in
$\mathbb Q \mathbb A_r$
.
Theorem 1.2 ([Reference McCullough and Pascoe12], Theorem 1.1)
An invertible operator T acting on a Hilbert space
$\mathcal {H}$
is in
$\mathbb {Q} \mathbb {A}_r$
if and only if there exists an invertible operator J acting on a larger Hilbert space
$\mathcal {K}$
containing
$\mathcal {H}$
such that
$T^n=P_{\mathcal {H}}J^n|_{\mathcal {H}}$
for all
$n \in \mathbb {Z}$
and J, up to unitary equivalence, takes the following form
$$\begin{align*}J=U\begin{bmatrix} rI_{\mathcal{K}_{0}} & 0\\ 0 & r^{-1}I_{\mathcal{K}_{1}} \end{bmatrix} \end{align*}$$
where U is a unitary,
${\mathcal {K}}={\mathcal {K}}_0 \oplus {\mathcal {K}}_1$
and
$P_{\mathcal {H}}$
denotes the orthogonal projection of
${\mathcal {K}}$
onto
${\mathcal {H}}$
.
These wider classes of operators generalizing the
$A_r$
-contractions have been extensively studied in recent past in [Reference Mittal14, Reference Tsikalas16, Reference Tsikalas17] also.
In this article, we further analyse the
$C_{1,r}$
class and the quantum annulus. We find the following model-cum-characterization theorem for the
$C_{1,r}$
class in terms of a pair of
$A_r$
-unitaries. Before we state the theorem, let us mention that any rational function f with poles off
$\overline {A}_r$
can be represented as
$$ \begin{align} f(z)=\frac{p(z)}{q_1(z)q_2(z)}, \end{align} $$
where
$p, q_1$
and
$q_2$
are polynomials in
$\mathbb {C}[z]$
such that the zeros of
$q_1$
and
$q_2$
lie in
$\mathbb {C}\setminus \overline {\mathbb {D}}$
and
$r\mathbb {D}$
respectively.
Theorem 1.3 Let T be an invertible operator acting on a Hilbert space
$\mathcal {H}$
. Then
$T \in C_{1, r}$
if and only if there is a Hilbert space
$\mathcal {K} \supseteq {\mathcal {H}}$
, an
$A_r$
-unitary N on
$\mathcal K$
and a self adjoint unitary
$F $
on
$\mathcal K$
such that
$$\begin{align*}f(T)=P_{\mathcal{H}} \bigg(p(N)q_1(N)^{-1}q_2(FNF)^{-1}\bigg) \bigg|_{\mathcal{H}} \end{align*}$$
for every rational function f with poles off
$\overline {A}_r (\text {as in} \ (1.2))$
.
Note that both N and
$FNF^{-1}$
as in the model above are
$A_r$
-unitaries. Next, we obtain the following characterizations for an operator in the
$C_{1,r}$
class. We mention that the equivalence of the conditions
$(1)$
and
$(2)$
of this theorem follows from Proposition 3.2 in [Reference McCullough and Pascoe12]. However, we present a different proof for this part in this article.
Theorem 1.4 Let T be an invertible operator acting on a Hilbert space
${\mathcal {H}}$
. Then the following are equivalent:
-
(1)
$T \in C_{1, r}$
; -
(2)
$(1+r^2)I_{\mathcal {H}}-T^*T-r^2T^{-1}(T^{-1})^* \geq 0$
; -
(3)
$(T^*T)^{1/ 2} \in C_\alpha $
; -
(4)
$(T^*T)^{1/ 2}$
is an
$A_r$
-contraction ; -
(5) there exist a unitary U and an
$A_r$
-contraction P on
${\mathcal {H}}$
such that
$T=UP$
.
These results will be proved in Section 3. Though, we have
$C_{1,r} \subsetneq \mathbb {Q} \mathbb {A}_r$
by definition, actually these two classes are comparable. To see this, let us consider the map
$ \varphi : \mathbb {A}_r \to A_{r^2}$
defined by
$ \varphi (z)=rz$
, which is a biholomorphism with
$\varphi ^{-1}(z)=r^{-1}z$
. Now one can easily prove the following lemma that establishes the equivalence of
$C_{1,r}$
and
$\mathbb {Q} \mathbb {A}_r$
.
Lemma 1.5 An operator
$T \in C_{1, r}$
if and only if
$r^{-1/ 2} T \in \mathbb {Q} \mathbb {A}_{\sqrt {r}} $
. Also,
$T \in \mathbb {Q} \mathbb {A}_r$
if and only if
$rT \in C_{1, r^2}$
.
Hence, any result that holds for the
$C_{1,r}$
class must have an analog for the quantum annulus. So, we have the following model theorem and characterizations for
$\mathbb {Q} \mathbb {A}_r$
that are analogous to Theorems 1.3 & 1.4 respectively.
Theorem 1.6 Let T be an invertible operator acting on a Hilbert space
$\mathcal {H}$
. Then
$T \in \mathbb {Q} \mathbb {A}_r$
if and only if there is a Hilbert space
$\mathcal {K} \supseteq {\mathcal {H}}$
, a normal operator N on
$\mathcal K$
with
$\sigma (N) \subseteq \partial \overline {\mathbb {A}}_r$
and a self adjoint unitary
$F $
on
$\mathcal K$
such that
$$\begin{align*}g(T)=P_{\mathcal{H}} \bigg(p(N)q_1(N)^{-1}q_2(FNF)^{-1}\bigg) \bigg|_{\mathcal{H}} \end{align*}$$
for every rational function
$g=p/ q_1q_2$
with the zeros of
$q_1$
and
$q_2$
in
$\mathbb {C} \setminus r^{-1} \overline {\mathbb {D}}$
and
$r\mathbb {D}$
respectively.
Theorem 1.7 Let T be an invertible operator on a Hilbert space
${\mathcal {H}}$
. Then the following are equivalent:
-
(1)
$T \in \mathbb {Q} \mathbb {A}_{r}$
; -
(2)
$(r^{-2}+r^2)I_{\mathcal {H}}-T^*T-T^{-1}(T^{-1})^* \geq 0$
; -
(3)
$\overline {\mathbb {A}}_r$
is a spectral set for
$(T^*T)^{1/ 2}$
; -
(4) there exist an operator P with
$\overline {\mathbb {A}}_r$
as a spectral set and a unitary U on
${\mathcal {H}}$
such that
$T=UP$
.
Note that the model that we obtain for
$\mathbb Q \mathbb A_r$
as in Theorem 1.6 consists of a pair of normal operators having their spectrums on the boundary of the annulus
$\mathbb {A}_r$
. Since Lemma 1.5 allows us to move back and forth between
$C_{1,r}$
and
$\mathbb {Q} \mathbb {A}_r$
, we do not want to miss the opportunity to achieve an alternative model for the
$C_{1,r}$
class which goes parallel with the model for
$\mathbb {Q} \mathbb {A}_r$
obtained by McCullough and Pascoe (i.e., Theorem 1.2).
Theorem 1.8 An invertible operator T acting on a Hilbert space
$\mathcal {H}$
is in
$C_{1, r}$
if and only if there is an invertible operator J on a Hilbert space
$\mathcal {K} \supseteq \mathcal {H}$
such that
$T^n=P_{\mathcal {H}}J^n|_{\mathcal {H}}$
for every
$n \in \mathbb {Z}$
and J admits the following form:
$$\begin{align*}J=U\begin{bmatrix} rI_{\mathcal{K}_0} & 0\\ 0 & I_{\mathcal{K}_1} \end{bmatrix}, \end{align*}$$
where U is a unitary and
$\mathcal {K}=\mathcal {K}_0 \oplus \mathcal {K}_1$
.
We prove the results associated with
$\mathbb {Q} \mathbb {A}_r$
in Section 4. In Section 2, we prove a few relevant and preparatory results.
2 Preparatory results
We begin with a famous result due to Ando which states that a pair of commuting contractions
$T_1, T_2$
can always be lifted simultaneously to a pair of commuting unitaries
$U_1,U_2$
.
Theorem 2.1 (Ando, [Reference Ando3])
Given a commuting pair of contractions
$(T_1, T_2)$
on a Hilbert space
$\mathcal {H},$
there exists a commuting pair of unitaries
$(U_1, U_2)$
on a Hilbert space
$\mathcal {K}$
containing
$\mathcal {H}$
such that
$$\begin{align*}p(T_1, T_2)=P_{\mathcal{H}}p(U_1, U_2)|_{\mathcal{H}} \end{align*}$$
for every polynomial p in two variables.
We now state and prove a few basic properties of an operator in
$C_{1,r}$
class which will be used in the proof of the main theorems.
Lemma 2.2 For every operator T in
$C_{1,r}$
class, we have
$r \leq \|T\| \leq 1.$
A proof to this result follows from the fact that
$\|T\|, \|rT^{-1}\| \leq 1$
. The converse of Lemma 2.2 does not hold. Indeed, if we choose
$r=1/2$
and
$ T =\begin {bmatrix} 0 & 1 \\ \frac {1}{100} & 0 \end {bmatrix}, $
then T is invertible and
$r \leq \|T\| \leq 1$
. However, T does not belong to
$C_{1,r}$
as the spectrum
$\sigma (T)$
is not contained in
$\overline {A}_r$
. Even more is true, the converse to Lemma 2.2 does not hold for
$A_r$
-contractions, i.e., an invertible operator T with
$\sigma (T) \subseteq \overline {A}_r$
and
$r \leq \|T\| \leq 1$
is not necessarily an
$A_r$
-contraction. The following example due to G. Misra [Reference Misra13] shows this clearly. Before going to the example let us state an interesting result due to Williams [Reference Williams18] showing an interplay between spectral set and complete non-normality. It is to mention that an operator
$T \in \mathcal {B}(\mathcal {H})$
is said to be completely non-normal if there is no nonzero closed subspace
${\mathcal {H}}_1$
of
${\mathcal {H}}$
that reduces T and
$T|_{{\mathcal {H}}_1}$
is normal.
Theorem 2.3 (Williams, [Reference Williams18])
If an operator T on a finite dimensional space is completely non-normal and
$\|T\|=1$
, then
$\overline {\mathbb {D}}$
is a minimal spectral set for T, i.e., no proper closed subset of
$\overline {\mathbb {D}}$
is a spectral set for T.
Example 2.4 For
$0<r<1,$
the matrix
$ T=\begin {bmatrix} \sqrt {r} & 1-r\\ 0 & \sqrt {r} \end {bmatrix} $
is invertible and
$\sigma (T)=\{\sqrt {r}\} \subseteq A_r.$
Note that
$$\begin{align*}T^*T=\begin{bmatrix} r & \sqrt{r}(1-r)\\ \sqrt{r}(1-r) & r+(1-r)^2\\ \end{bmatrix} \end{align*}$$
and thus it follows that
$\sigma (T^*T)=\{1, r^2\}.$
Therefore,
$ \|T\|^2=\|T^*T\|=1. $
Hence,
$r\leq \|T\| \leq 1$
. It is not difficult to see that
$T \in C_{1, r}$
. Since
$\|T\|=1$
and T is completely non-normal, Theorem 2.3 implies that
$\overline {\mathbb {D}}$
is a minimal spectral set for
$T.$
Hence
$\overline {A}_r$
cannot be a spectral set for T.
Lemma 2.5 Let T be an invertible operator acting on a Hilbert space
${\mathcal {H}}$
. Then
-
(a) T is an
$A_r$
-contraction if and only if
$rT^{-1}$
is an
$A_r$
-contraction ; -
(b) T is in
$C_{1,r}$
if and only if
$rT^{-1}$
is in
$C_{1,r}$
.
Proof The part-(b) is obvious and so we prove only part-(a). Let T be an
$A_r$
-contraction. Then by Lemma 2.2, we have
$\sigma (T)\subseteq \overline {A}_r$
and
$r \leq \|T\| \leq 1$
. So, it follows from the spectral mapping theorem that
$$\begin{align*}\sigma(rT^{-1})=\{ r/ \lambda \ :\ \lambda \in \sigma(T) \} \subseteq \overline{A}_r. \end{align*}$$
Let f be any rational function with poles off
$\overline {A}_r.$
Then we define a rational function
$g(z)=f(rz^{-1})$
, where
$z \mapsto rz^{-1}$
is an automorphism of the annulus
$A_r$
. Evidently, g has its poles off
$\overline {A}_r.$
Now,
$$ \begin{align*} \|f(rT^{-1})\| &=\|g(T)\| \leq \sup \{|g(z)| : r \leq |z| \leq 1\} \\& = |g(w)| \ \ \ \ \text{for some } w \in \mathbb{T} \cup r\mathbb{T}\ \ \ \ \text{(by the Maximum-modulus principle)}\\&=|f(rw^{-1})| \\& \leq \sup \{|f(z)| : r \leq |z| \leq 1\}. \\ \end{align*} $$
Therefore,
$\overline {A}_r$
is a spectral set for
$rT^{-1}$
. Again if
$\overline {A}_r$
is a spectral set for
$S=rT^{-1}$
, then by previous part of the proof we have that
$rS^{-1}=T$
is also an
$A_r$
-contraction and the proof is complete.
Let T be an operator on a Hilbert space
${\mathcal {H}}$
and let
$\gamma $
be a simple closed curve in
$\mathbb {C}$
such that
$\sigma (T)$
is contained in the interior of
$\gamma $
. If f is a holomorphic function on and in the interior of
$\gamma $
, then
$f(T)$
can be defined in the following way:
$$ \begin{align} f(T):=\frac{1}{2\pi i}\underset{\gamma}{\int}f(w)(w-T)^{-1}dw. \end{align} $$
It is merely mentioned that the above integral and hence the definition of
$f(T)$
is independent of the choice of
$\gamma $
. Before going to the next proposition we state a classic result whose proof is a routine exercise.
Lemma 2.6 Let
$T \in \mathcal {B}(\mathcal {H})$
and let
$\Omega $
be an open set containing
$\sigma (T).$
If a sequence
$\{f_n\}$
of holomorphic functions on
$\Omega $
converges uniformly to a function f on every compact subset of
$\Omega $
, then f is holomorphic on
$\Omega $
and
$\{f_n(T)\}$
as in
$($
2.1
$)$
converges to
$f(T)$
in operator norm.
Every rational function f with poles off
$\overline {A}_r$
is analytic in an open neighborhood containing
$\overline {A}_r$
and thus has a unique Laurent series
$f(z)=\overset {\infty }{\underset {j=-\infty }{\sum }}f_jz^j$
. We show that for
$T \in C_{1, r}$
acting on
$\mathcal {H},$
the series
$\overset {\infty }{\underset {j=-\infty }{\sum }}f_jT^j$
defines an operator on
$\mathcal {H}$
and
$f(T)=p(T)q(T)^{-1}=\overset {\infty }{\underset {j=-\infty }{\sum }}f_jT^j$
, where
$f=p/q$
with q having no zeros inside
$\overline {A}_r$
.
Proposition 2.7 Given an operator
$T \in C_{1, r}$
on a Hilbert space
$\mathcal {H}$
and a rational function f with poles off
$\overline {A}_r,$
the series
$\overset {\infty }{\underset {j=-\infty }{\sum }}f_jT^j$
defines a bounded linear operator on
$\mathcal {H}$
and is same as
$f(T)$
.
Proof Since
$T \in C_{1, r}$
, it follows from the spectral mapping theorem that
$\sigma (T) \subseteq \overline {A}_r$
. Thus
$f(T)$
is well-defined. The sequence
$\tilde {f}_n(z)=\overset {n}{\underset {j=-n}{\sum }}f_jz^j$
converges uniformly to
$f(z)$
on
$\overline {A}_r$
and we have by Lemma 2.6 that
$\tilde {f}_n(T)$
converges to
$f(T)$
in operator norm topology. Thus,
$f(T)=\overset {\infty }{\underset {j=-\infty }{\sum }}f_jT^j.$
In order to prove that an
$A_r$
-contraction
$T \in \mathcal B({\mathcal {H}})$
admits an
$A_r$
-unitary dilation N on
$\mathcal K \supseteq {\mathcal {H}}$
, one needs to show that
$f(T)=P_{\mathcal {H}}f(N)|_{\mathcal {H}}$
for every rational function f with poles off
$\overline {A}_r$
. The next lemma shows that instead of all rational functions f it suffices to consider only the integral powers of z, i.e., the functions of the type
$z^j$
for
$j \in \mathbb Z$
.
Lemma 2.8 For operators T and N in
$C_{1, r}$
acting on Hilbert spaces
${\mathcal {H}}$
and
$\mathcal {K}$
respectively with
$\mathcal {K}\supseteq \mathcal {H}$
, the following are equivalent:
-
(1)
$f(T)=P_{\mathcal {H}}f(N)|_{\mathcal {H}}$
for every rational function f with poles off
$\overline {A}_r$
; -
(2)
$T^j=P_{\mathcal {H}}N^j|_{\mathcal {H}}$
for every
$j \in \mathbb {Z}.$
Proof
$(1) \implies (2)$
is obvious. We prove
$(2) \implies (1)$
. Let f be a rational function with poles off
$\overline {A}_r$
and let
$f(z)=\overset {\infty }{\underset {j=-\infty }{\sum }}f_jz^j$
be its Laurent series. We have by Proposition 2.7 that
$f(T)=\overset {\infty }{\underset {j=-\infty }{\sum }}f_jT^j$
and
$f(N)=\overset {\infty }{\underset {j=-\infty }{\sum }}f_jN^j$
. For any
$h\in {\mathcal {H}}$
, it follows that
$$\begin{align*}f(T)h =\overset{\infty}{\underset{j=-\infty}{\sum}}f_jT^jh =\overset{\infty}{\underset{j=-\infty}{\sum}}f_j(P_{\mathcal{H}}N^jh) =P_{\mathcal{H}}\overset{\infty}{\underset{j=-\infty}{\sum}}f_jN^jh =P_{\mathcal{H}}f(N)h. \end{align*}$$
Next, we show that the three classes
$\mathcal A_r, C_\alpha , C_{1,r}$
as in Theorem 1.1 agree under subnormality condition. To do so, it suffices to show that every subnormal
$C_{1, r}$
operator is an
$A_r$
-contraction. First we state an elementary result whose proof is a routine exercise.
Lemma 2.9 Let N be a normal operator with
$\sigma (N) \subseteq \overline {A}_r$
. Then N is an
$A_r$
-contraction.
The next two results are also important in the context of this article.
Proposition 2.10 A subnormal operator
$T \in C_{1, r}$
if and only if T is an
$A_r$
-contraction.
Proof Let
$T \in C_{1, r}$
be subnormal. We have that
$\sigma (T) \subseteq \overline {A}_r$
. Let N be the minimal normal extension of T. It follows from [Reference Conway5, Chapter 2, Theorem 2.11] that
$\sigma (N) \subseteq \sigma (T)$
and so,
$\sigma (N) \subseteq \overline {A}_r$
. Note that N is invertible and since
$T=N|_{\mathcal {H}}$
, we have that
$T^{-1}=N^{-1}|_{\mathcal {H}}$
. Therefore,
$T^m=N^m|_{\mathcal {H}}$
for every
$m \in \mathbb Z$
. It follows from Lemma 2.8 that
$f(T)=f(N)|_{\mathcal {H}}$
for every rational function f with poles outside
$\overline {A}_r$
. By Lemma 2.9, N is an
$A_r$
-contraction and so,
$\|f(T)\| \leq \|f(N)\| \leq \sup \{|f(z)| : z \in \overline {A}_r\}$
for every rational function f with poles off
$\overline {A}_r$
. Thus, T is an
$A_r$
-contraction. The converse follows from Theorem 1.1.
Proposition 2.11 A subnormal operator
$T \in \mathbb {Q} \mathbb {A}_r$
if and only if
$\overline {\mathbb {A}}_r$
is a spectral set for T.
Proof Let
$T \in \mathbb {Q}\mathbb {A}_r$
be subnormal. We have by Lemma 1.5 that
$rT \in C_{1, r^2}$
is also subnormal. It follows from Proposition 2.10 that
$rT$
is an
$A_{r^2}$
-contraction. Since
$\varphi : \mathbb {A}_r \to A_{r^2}, \varphi (z)=rz$
is a biholomorphism, we have that
$\varphi ^{-1}(rT)=T$
has
$\varphi ^{-1}(\overline {A}_{r^2})=\overline {\mathbb {A}}_r$
as a spectral set. The converse is trivial.
However, an operator in
$\mathbb Q \mathbb A_r$
may not always have
$\overline {\mathbb A}_r$
as a spectral set. Actually, the class of operators having
$\overline {\mathbb A}_r$
as a spectral set is contained in
$\mathbb Q \mathbb A_r$
and it follows trivially from the von Neumann’s inequality. The following example shows that the containment is strict.
Example 2.12 For
$0<r<1$
, consider the matrix
$ T=\begin {bmatrix} 1 & -r+r^{-1}\\ 0 & 1 \end {bmatrix} $
. It is not difficult to see that T is invertible and
$\|T\|=\|T^{-1}\| =r^{-1}$
. Therefore,
$T \in \mathbb {Q} \mathbb {A}_r$
. Since
$\|rT\|=1$
and
$rT$
is completely non-normal, Theorem 2.3 implies that
$\overline {\mathbb {D}}$
is a minimal spectral set for
$rT$
. Consequently,
$r^{-1}\overline {\mathbb {D}}$
is a minimal spectral set for T. Hence,
$\overline {\mathbb A}_r$
cannot be a spectral set for T.
Lemma 2.13 Let
$T \in C_{1, r}$
. If f as in
$($
1.2
$)$
is any rational function with poles off
$\overline {A}_r$
, then
$$\begin{align*}f(z)=\bigg(\overset{\infty}{\underset{n=0}{\sum}}a_{n}z^n\bigg) \bigg(\overset{\infty}{\underset{n=0}{\sum}}b_{n}z^{-n}\bigg) \; \text{ and } \; f(T)=\bigg(\overset{\infty}{\underset{n=0}{\sum}}a_{n}T^n\bigg) \bigg(\overset{\infty}{\underset{n=0}{\sum}}b_{n}T^{-n}\bigg), \end{align*}$$
for some scalar coefficients
$a_n, b_n$
.
Proof For any rational function f with poles off
$\overline {A}_r$
we have from (1.2) that
$$\begin{align*}f(z)=\frac{p(z)}{q_1(z)q_2(z)}, \end{align*}$$
where
$p, q_1$
and
$q_2$
are polynomials with zeros of
$q_1$
in
$\mathbb {C}\setminus \overline {\mathbb {D}}$
and zeros of
$q_2$
in
$r\mathbb {D}.$
Suppose
$\alpha _1, \dots , \alpha _k \in \mathbb C \setminus \overline {\mathbb D}$
are the zeros of
$q_1$
and
$\beta _1 ,\dots , \beta _l \in r \mathbb D$
are the zeros of
$q_2$
. Then
$$\begin{align*}q_1(z) \hspace{-0.5pt}=\hspace{-0.5pt}\alpha (z-\alpha_1)(z-\alpha_2)\dotsc (z-\alpha_k)\hspace{-0.6pt} \quad \text{and} \hspace{-0.6pt}\quad q_2(z) \hspace{-0.6pt}=\hspace{-0.6pt}\beta(z-\beta_1)(z-\beta_2)\dotsc (z-\beta_l), \end{align*}$$
for some
$\alpha , \beta \in \mathbb C$
. Now for each
$\alpha _j,\beta _i$
and for any
$z \in \overline {A}_r$
, we have
$$\begin{align*}\frac{1}{z-\alpha_j} = \frac{-1}{\alpha_j(1-z/ \alpha_j)} =-\overset{\infty}{\underset{n=0}{\sum}}\frac{z^n}{\alpha_j^{n+1}} \quad \text{and} \quad \frac{1}{z-\beta_i} =\frac{1}{z(1-\beta_i / z)} =\overset{\infty}{\underset{n=0}{\sum}}\beta_i^nz^{-(n+1)}, \end{align*}$$
where both the series converge uniformly on
$\overline {A}_r.$
Thus, for any
$z\in \overline {A}_r$
we have that
$$\begin{align*}\frac{1}{q_1(z)}=\overset{\infty}{\underset{n=0}{\sum}}q_{n1}z^n \quad \text{and} \quad \frac{1}{q_2(z)}=\overset{\infty}{\underset{n=0}{\sum}}\frac{q_{n2}}{z^n} \end{align*}$$
for some scalar coefficients
$q_{n1}, q_{n2}$
. Consequently,
$$\begin{align*}f(z)=p(z)\bigg(\overset{\infty}{\underset{n=0}{\sum}}q_{n1}z^n\bigg) \bigg(\overset{\infty}{\underset{n=0}{\sum}}q_{n2}z^{-n}\bigg). \end{align*}$$
Evidently it follows from Lemma 2.6 that
$$\begin{align*}f(T)=p(T)\bigg(\overset{\infty}{\underset{n=0}{\sum}}q_{n1}T^n\bigg) \bigg(\overset{\infty}{\underset{n=0}{\sum}}q_{n2}T^{-n}\bigg) \end{align*}$$
and the proof is complete.
We conclude this Section by recalling from the literature a useful result on joint spectrum.
Theorem 2.14 ([Reference Taylor15], Theorem 4.9)
If
$\underline {T}=(T_1, \dotsc , T_n)$
is a tuple of commuting operators on a Hilbert space X and if
$\sigma _T(\underline {T})=K_1 \cup K_2$
, where
$K_1$
and
$K_2$
are disjoint compact sets in
$\mathbb {C}^n,$
then there are closed linear subspaces
$X_1$
and
$X_2$
of X such that
-
(1)
$X=X_1\oplus X_2;$
-
(2)
$X_1, X_2$
are invariant under any operator which commutes with each
$T_k;$
-
(3)
$\sigma _T(\underline {T}|_{X_1})=K_1$
and
$\sigma _T(\underline {T}|_{X_2})=K_2$
, where
$\underline {T}|_{X_i}=(T_1|_{X_i}, \dotsc , T_n|_{X_i})$
for
$i=1,2.$
3 Characterizations and operator model for the
$C_{1, r}$
class
While investigating the success or failure of rational dilation on the closure of a domain
$\Omega $
, a primary step toward the endeavor is to study the normal operators having their spectrum in the boundary
$\partial \overline {\Omega }$
. Such operators constitute an analog of unitaries, which are normal operators associated with the boundary of the unit disk. For an annulus
$A_r$
, they are
$A_r$
-unitaries. In this section, we first characterize an
$A_r$
-unitary as a direct sum
$U_1 \oplus rU_2$
for a pair of unitaries
$U_1,U_2$
. Using this characterization, we frame a model (see Theorem 1.3) for an operator in
$C_{1, r}$
class and the model consists of a pair of
$A_r$
-unitaries. For the sake of brevity, we fix the following notations for a contraction
$T \in \mathcal {B}({\mathcal {H}})$
:
$$\begin{align*}T(n)=T^n \ \ (n \geq 1), \quad T(0)=I_{\mathcal{H}}, \quad T(n)=T^{*|n|} \ \ (n \leq -1). \end{align*}$$
Evidently,
$\|T\| \leq 1$
if and only if
$I-T^*T \geq 0$
. Let
$D_T=(I-T^*T)^{1/ 2}$
be the unique positive square root of a contraction T.
Theorem 3.1 An operator
$T \in \mathcal {B}(\mathcal {H})$
is an
$A_r$
-unitary if and only if
$\mathcal {H}$
decomposes into an orthogonal sum
$\mathcal {H}=\mathcal {H}_1\oplus \mathcal {H}_2$
such that
$\mathcal {H}_1, \mathcal {H}_2$
reduce T and
$T_1= T|_{\mathcal {H}_1}, T_2=rT^{-1}|_{\mathcal {H}_2}$
are unitaries. This decomposition is uniquely determined. Indeed, we have that
$$ \begin{align*} \mathcal{H}_1=\{h \in \mathcal{H} \ : \ \|T^nh\|=\|h\|=\|T^{*n}h\|, \;\; n=1,2,\dotsc \} \end{align*} $$
and
$$ \begin{align*} \mathcal{H}_2=\{h \in \mathcal{H} \ : \ \|T^nh\|=r^n\|h\|=\|T^{*n}h\|, \;\; \ n=-1,-2,\dotsc \}. \end{align*} $$
The space
$\mathcal {H}_1$
or
$\mathcal {H}_2$
may coincide with the trivial space
$\{0\}$
. With respect to the decomposition
$\mathcal {H}=\mathcal {H}_1\oplus \mathcal {H}_2$
, T has the following block-matrix form:
$$\begin{align*}T=\begin{bmatrix} T_1 & 0 \\ 0 & rT_2^{-1} \\ \end{bmatrix}. \end{align*}$$
Proof Since T is an
$A_r$
-unitary, we have by Theorem 1.1 that T is in
$C_{1,r}$
and thus T and
$rT^{-1}$
are contractions. Hence,
$T(n)$
and
$(rT^{-1})(n)$
are contractions for every integer n. For each fixed
$n \in \mathbb {Z}$
, it is evident that
$$\begin{align*}& \text{Ker}\,D_{T(n)} =\{h\in {\mathcal{H}} \,:\, \|T(n)h\|=\|h\| \} \quad \text{and} \quad \\ &\text{Ker}\,D_{(rT^{-1})(n)} =\{h\in {\mathcal{H}} \,:\, \|rT^{-1}(n)h\|=\|h\| \}. \end{align*}$$
Therefore, we have
$$\begin{align*}\mathcal{H}_1=\underset{n \in \mathbb{Z} \setminus \{0\}}{\bigcap}\text{Ker}\ D_{T(n)} \quad \text{and} \quad\mathcal{H}_2=\underset{n \in \mathbb{Z} \setminus \{0\}}{\bigcap}\text{Ker}\ D_{(rT^{-1})(n)}. \end{align*}$$
It is obvious that
$\mathcal {H}_1$
and
$\mathcal {H}_2$
are closed linear subspaces of
$\mathcal {H}$
. For any
$h \in \mathcal {H}_1$
, we have
$$ \begin{align*} & \|T^nTh\|=\|T^{n+1}h\|=\|h\|=\|Th\| \ \ (n=0,1,2, \dotsc), \\ \|T^{*n}Th\| &=\|T^{*n-1}T^*Th\|=\|T^{*n-1}h\|=\|h\|=\|Th\| \ \ (n=1,2, \dotsc), \end{align*} $$
which follow from the fact that for a contraction
$T, \|Th\|=\|h\|$
if and only if
$T^*Th=h$
. Hence,
$Th \in \mathcal {H}_1$
. Similarly, one can show that
$T^*h \in \mathcal {H}_1$
. Thus
$\mathcal {H}_1$
reduces T. A similar argument implies that
$\mathcal {H}_2$
reduces T. If we set
$T_1=T|_{\mathcal {H}_1}$
and
$T_2=rT^{-1}|_{\mathcal {H}_2},$
then it follows from the definition of
${\mathcal {H}}_1$
and
${\mathcal {H}}_2$
that
$T_1$
and
$T_2$
are unitaries on
${\mathcal {H}}_1$
and
${\mathcal {H}}_2$
respectively. Consequently, we have that
$$ \begin{align*} \langle h_1, h_2\rangle=\langle T^{*}Th_1, (rT^{-1})(rT^{-1})^*h_2\rangle=\langle T^{-1}T^{-1*}T^*Th_1, r^2h_2 \rangle=r^2 \langle h_1,h_2\rangle \end{align*} $$
for
$h_1 \in {\mathcal {H}}_1, h_2 \in {\mathcal {H}}_2$
and so,
$\langle h_1, h_2 \rangle =0$
as
$0<r<1$
. Hence,
$\mathcal {H}_1$
and
$\mathcal {H}_2$
are orthogonal. Consider the subspace
$\mathcal {H}_3=\mathcal {H} \ominus \big (\mathcal {H}_1 \oplus \mathcal {H}_2\big )$
which reduces T and thus,
$T_3=T|_{\mathcal {H}_3}$
is normal. Since
$\sigma (T_3) \subseteq \sigma (T) \subseteq \mathbb {T} \cup r\mathbb {T}$
, we have that
$\sigma (T_3)=K_1 \cup K_2,$
where
$K_1=\sigma (T_3)\cap \mathbb {T}$
and
$K_2=\sigma (T_3) \cap r\mathbb {T}$
. Since
$T_3$
is normal, we have by Theorem 2.14 that there are closed subspaces
$\mathcal {H}_{3}'$
and
$\mathcal {H}_3"$
of
${\mathcal {H}}_3$
reducing
$T_3$
such that
$$\begin{align*}\mathcal{H}_{3}=\mathcal{H}_{3}' \oplus \mathcal{H}_{3}", \quad \sigma(T|_{\mathcal{H}_3'})=K_1 \subseteq \mathbb{T} \quad \text{and} \quad \sigma(T|_{\mathcal{H}_3"})=K_2 \subseteq r\mathbb{T}. \end{align*}$$
It shows that T and
$rT^{-1}$
are unitaries on
$\mathcal {H}_3'$
and
$\mathcal {H}_3"$
respectively. Thus,
$\mathcal {H}_3'\subseteq \mathcal {H}_1$
and
$\mathcal {H}_3" \subseteq \mathcal {H}_2$
implying that
$\mathcal {H}_3=\mathcal {H}_3' \oplus \mathcal {H}_3" \subseteq \mathcal {H}_1 \oplus \mathcal {H}_2$
. Consequently,
$\mathcal {H}_3=\{0\}$
and so,
$\mathcal {H}=\mathcal {H}_1 \oplus \mathcal {H}_2$
. Also, it is clear from the construction that
${\mathcal {H}}_1$
and
${\mathcal {H}}_2$
are the maximal closed reducing subspaces of
${\mathcal {H}}$
on which T and
$rT^{-1}$
act as unitaries respectively. Hence,
${\mathcal {H}}_1, {\mathcal {H}}_2$
are uniquely determined and the proof is complete.
Now we are in a position to give a proof to Theorem 1.3, one of the main results of this article.
Proof of Theorem 1.3.
Let
$T \in C_{1, r}$
. Then
$(T, rT^{-1})$
is a commuting pair of contractions acting on
$\mathcal {H}$
. It follows from Ando’s dilation, Theorem 2.1, that there are commuting unitaries
$U_1, rU_2^{-1}$
on a Hilbert space
$\mathcal {K}_0\supseteq \mathcal {H}$
such that
$$ \begin{align} p(T, rT^{-1})h=P_{\mathcal{H}}p(U_1, rU_2^{-1})h \end{align} $$
for every
$h \in \mathcal {H}$
and for every polynomial
$p \in \mathbb {C}[z_1,z_2]$
. Consequently, we have that
$$ \begin{align} T^jh =P_{\mathcal{H}}U_1^jh \quad \text{and} \quad \ T^{-j}h =P_{\mathcal{H}}U_2^{-j}h, \quad \text{for} \ h \in \mathcal{H} \ \text{and} \ j=0, 1, 2, \dotsc \,. \end{align} $$
Let f be a rational function with poles off
$\overline {A}_r$
. Then,
$f(z)=p(z)q_1(z)^{-1}q_2(z)^{-1}$
as in (1.2), where
$q_1, q_2$
have their zeros in
$\mathbb {C} \setminus \overline {\mathbb {D}}$
and
$r\mathbb {D}$
respectively. For any
$h \in {\mathcal {H}}$
, we have from Lemma 2.13 that
$$ \begin{align*} f(T)h&=p(T)\bigg(\overset{\infty}{\underset{n=0}{\sum}}q_{n1}T^n\bigg) \bigg(\overset{\infty}{\underset{n=0}{\sum}}q_{n2}T^{-n}\bigg)h\\ &=\lim_{m \to \infty} \bigg[ p(T)\bigg(\overset{m}{\underset{n=0}{\sum}}q_{n1}T^n\bigg) \bigg(\overset{m}{\underset{n=0}{\sum}}q_{n2}T^{-n}\bigg)\bigg]h \\ &=\lim_{m \to \infty} \bigg[ P_{\mathcal{H}}p(U_1)\bigg(\overset{m}{\underset{n=0}{\sum}}q_{n1}U_1^n\bigg) \bigg(\overset{m}{\underset{n=0}{\sum}}q_{n2}U_2^{-n}\bigg)\bigg]h \quad [\text{by } (3.1) \; \& \; (3.2)]\\ &=P_{\mathcal{H}}p(U_1)\bigg(\overset{\infty}{\underset{n=0}{\sum}}q_{n1}U_1^n\bigg) \bigg(\overset{\infty}{\underset{n=0}{\sum}}q_{n2}U_2^{-n}\bigg)h\\ &=P_{\mathcal{H}}p(U_1)q_1(U_1)^{-1}q_2(U_2)^{-1}h.\\ \end{align*} $$
Set
$$\begin{align*}N=\begin{bmatrix} U_1 & 0\\ 0 & U_2\\ \end{bmatrix} \ \text{and} \ \ F=\begin{bmatrix} 0 & I_{\mathcal{K}_0}\\ I_{\mathcal{K}_0} & 0\\ \end{bmatrix} \quad \text{ on } \; \mathcal{K}=\mathcal{K}_0 \oplus \mathcal{K}_0. \end{align*}$$
Note that F is a self-adjoint unitary. Since
$U_1$
and
$r^{-1}U_2$
are unitaries on
$\mathcal {K}_0$
, it follows from Theorem 3.1 that N is an
$A_r$
-unitary. We have that
${\mathcal {H}} \subseteq \mathcal K_0$
. Let
$V: \mathcal {H} \to \mathcal {K}$
be defined as
$Vh=(h, 0)$
. Evidently, V is an isometric embedding and
$V^*(x_1, x_2)=P_{\mathcal {H}}x_1$
for every
$(x_1, x_2) \in \mathcal {K}$
. So, for any
$h\in {\mathcal {H}}$
we have that
$$ \begin{align*} & V^*\bigg(p(N)q_1(N)^{-1}Fq_2(N)^{-1}F\bigg)Vh \\ &\quad = V^*\bigg(p(N)q_1(N)^{-1}Fq_2(N)^{-1}\begin{bmatrix} 0 & I_{\mathcal{K}_0}\\ I_{\mathcal{K}_0} & 0\\ \end{bmatrix}\bigg) \begin{bmatrix} h \\ 0 \\ \end{bmatrix}\\ &\quad =V^*\bigg(p(N)q_1(N)^{-1}F\begin{bmatrix} q_2(U_1)^{-1} & 0\\ 0 & q_2(U_2)^{-1}\\ \end{bmatrix}\bigg) \begin{bmatrix} 0 \\ h \\ \end{bmatrix}\\ &\quad =V^*\bigg(p(N)q_1(N)^{-1}\begin{bmatrix} 0 & I_{\mathcal{K}_0}\\ I_{\mathcal{K}_0} & 0\\ \end{bmatrix}\bigg) \begin{bmatrix} 0 \\ q_2(U_2)^{-1}h \\ \end{bmatrix}\\ &\quad =V^*\begin{bmatrix} p(U_1)q_1(U_1)^{-1} & 0\\ 0 & p(U_2)q_1(U_2)^{-1}\\ \end{bmatrix} \begin{bmatrix} q_2(U_2)^{-1}h \\ 0\\ \end{bmatrix}\\ &\quad =V^*\begin{bmatrix} p(U_1)q_1(U_1)^{-1}q_2(U_2)^{-1}h \\ 0\\ \end{bmatrix}\\ &\quad =P_{\mathcal{H}}p(U_1)q_1(U_1)^{-1}q_2(U_2)^{-1}h\\ &\quad =f(T)h. \end{align*} $$
Also, we have
$$\begin{align*}FNF=\begin{bmatrix} 0 & I_{\mathcal{K}_0}\\ I_{\mathcal{K}_0} & 0\\ \end{bmatrix} \begin{bmatrix} U_1 & 0\\ 0 & U_2\\ \end{bmatrix} \begin{bmatrix} 0 & I_{\mathcal{K}_0}\\ I_{\mathcal{K}_0} & 0\\ \end{bmatrix} = \begin{bmatrix} U_2 & 0\\ 0 & U_1\\ \end{bmatrix} \end{align*}$$
and hence
$q_2(FNF)=Fq_2(N)F$
. Note that
$FNF$
is unitarily equivalent to N and hence is an
$A_r$
-unitary. Since F is a self-adjoint unitary, we have that
$q_2(FNF)^{-1}=Fq_2(N)^{-1}F$
. Putting everything together, we have that
$$\begin{align*}f(T)h=V^*\bigg(p(N)q_1(N)^{-1}q_2(FNF)^{-1}\bigg)Vh \end{align*}$$
for any rational function f with poles off
$\overline {A}_r$
and for every
$h \in \mathcal {H}$
. To see the converse, assume that there is a Hilbert space
$\mathcal {K} \supseteq {\mathcal {H}}$
, an
$A_r$
-unitary N on
$\mathcal K$
and a self adjoint unitary
$F $
on
$\mathcal K$
such that
$$\begin{align*}f(T)=P_{\mathcal{H}} \bigg(p(N)q_1(N)^{-1}q_2(FNF)^{-1}\bigg) \bigg|_{\mathcal{H}} \end{align*}$$
for every rational function f with poles off
$\overline {A}_r (\text {as in} \ (1.2))$
. Then
$T=P_{\mathcal {H}} N|_{\mathcal {H}}$
and
$T^{-1}=P_{\mathcal {H}}FN^{-1}F|_{\mathcal {H}}$
. Since N is an
$A_r$
-unitary, it follows from Theorem 1.1 that
$N \in C_{1, r}$
. Thus
$\|T\| \leq \|N\| \leq 1$
and
$\|T^{-1}\| \leq \|FN^{-1}F\| \leq \|N^{-1}\| \leq r^{-1}$
. Consequently,
$T \in C_{1, r}$
. The proof is now complete.
Remark 3.2 In Theorem 1.3, if we denote the
$A_r$
-unitary
$FNF$
by
$\widetilde {N}$
, then it follows as a special case that for every
$A_r$
-contraction
$T \in \mathcal B({\mathcal {H}})$
, there is a Hilbert space
$\mathcal K \supseteq {\mathcal {H}}$
and an
$A_r$
-unitary
$\widetilde {N} \in \mathcal B(\mathcal K)$
such that
$$\begin{align*}f(T)=P_{{\mathcal{H}}}f(\widetilde{N})|_{{\mathcal{H}}} \end{align*}$$
for every rational function of the form
$f=1/q$
such that the zeros of q lie inside
$r\mathbb {D}$
. Also, if
$g=p/q_1$
with
$q_1$
having its zeros inside
$\mathbb {C} \setminus \overline {\mathbb {D}}$
, then also we have
$$\begin{align*}g(T)= P_{{\mathcal{H}}}g({N})|_{{\mathcal{H}}}, \end{align*}$$
for some
$A_r$
-unitary
$N \in \mathcal B(\mathcal K_1)$
such that
$\mathcal K_1 \supseteq {\mathcal {H}} $
.
We conclude this Section by providing Theorem 1.4, another main theorem of this article that characterizes an operator in
$C_{1, r}$
in different ways. Once again we mention that the equivalence of parts
$(1)$
&
$(2)$
of this theorem follows from [Reference McCullough and Pascoe12, Proposition 3.2]. However, we present here a different proof for this part also.
Proof of Theorem 1.4.
$(1) \implies (2)$
. Let
$T \in C_{1, r}$
. It is easy to see that
$$ \begin{align} D_T^2D_{(rT^{-1})^*}^2=D_{(rT^{-1})^*}^2D_T^2=(1+r^2)I_{\mathcal{H}}-T^*T-r^2T^{-1}(T^{-1})^*. \end{align} $$
Consequently,
$p(D_T^2)p(D_{(rT^{-1})^*}^2)=p(D_{(rT^{-1})^*}^2)p(D_T^2)$
for every polynomial
$p\in \mathbb {C}[z]$
. Choose a sequence of polynomials
$p_n(x)$
that converges uniformly to
$x^{1/ 2}$
on the interval
$0 \leq x \leq 1$
. It follows from the spectral theorem that the sequence of operators
$p_n(B)$
converges to
$B^{1/ 2}$
for any positive operator B such that
$0 \leq B \leq I_{\mathcal {H}}$
. Applying (3.3) to these polynomials and taking the limit as
$n \to \infty $
, we have
$$\begin{align*}D_TD_{(rT^{-1})^*}=D_{(rT^{-1})^*}D_T. \end{align*}$$
Thus, for any
$h \in {\mathcal {H}}$
, we have that
$$\begin{align*}\left\langle\left((1+r^2)I_{\mathcal{H}}-T^*T-r^2T^{-1}(T^{-1})^*\right)h, h \right\rangle &=\left\langle D_T^2D_{(rT^{-1})^*}^2h, h \right\rangle \\ &=\|D_TD_{(rT^{-1})^*}h\|^2 \geq 0. \end{align*}$$
$(2) \implies (3).$
Let
$\Delta _T=(1+r^2)I_{\mathcal {H}}-T^*T-r^2T^{-1}(T^{-1})^*\geq 0$
and let
$P=(T^*T)^{1/ 2}$
. Note that P is invertible as T is invertible. Moreover, we have
$$ \begin{align*} 0 \leq P^*\Delta_T P=P((1+r^2)I_{\mathcal{H}}-P^2-r^2P^{-2})P=(I_{\mathcal{H}}-P^2)(P^2-r^2I_{\mathcal{H}})=\alpha(P^*, P). \end{align*} $$
$(3) \implies (4).$
Let
$P=(T^*T)^{1/ 2} \in C_\alpha $
. Then
$\alpha (P^*, P)=(I_{\mathcal {H}}-P^2)(P^2-r^2I_{\mathcal {H}}) \geq 0$
. Let
$\lambda \in \sigma (P)$
. It follows from the spectral theorem that
$(1-\lambda ^2)(\lambda ^2-r^2) \geq 0$
and this holds if and only if
$r \leq \lambda \leq 1$
. Therefore,
$\sigma (P) \subseteq \overline {A}_r$
. Consequently, we have by Lemma 2.9 that P is an
$A_r$
-contraction.
$(4) \implies (5).$
Let
$P=(T^*T)^{1/ 2}$
be an
$A_r$
-contraction. For
$U=TP^{-1}$
, we have that
$$\begin{align*}U^*U=P^{-1}T^*TP^{-1}=P^{-1}P^2P^{-1}=I_{\mathcal{H}} \quad \text{and} \quad UU^*=TP^{-2}T^*=T(T^*T)^{-1}T^*=I_{\mathcal{H}}. \end{align*}$$
Hence, U is a unitary on
${\mathcal {H}}$
and
$T=UP$
.
$(5) \implies (1).$
Let
$T=UP$
for a unitary U and an
$A_r$
-contraction P on
${\mathcal {H}}$
. Then
$rT^{-1}=(rP^{-1})U^*$
. Consequently, we have that
$\|T\| \leq \|P\| \leq 1$
and
$\|rT^{-1}\| \leq \|rP^{-1}\| \leq 1$
. The proof is now complete.
4 The quantum annulus
In this Section, we provide a model for operators in the quantum annulus
$\mathbb {Q} \mathbb {A}_r$
. Recall that
$$\begin{align*}\mathbb{Q} \mathbb{A}_r=\{T: T \ \text{is invertible and} \ \|rT\|, \|rT^{-1}\| \leq 1\}, \end{align*}$$
which is the quantization of the closed annulus
$\overline {\mathbb {A}}_r=\{z \in \mathbb {C}: z \ne 0 \ \text {and} \ |rz|, |rz^{-1}|\leq 1\}$
in the sense that the scalars in the annulus are replaced by operators with similar norm-bounds. With this terminology,
$C_{1, r}$
is nothing but the quantization of
$\overline {A}_r$
. It is evident that
$C_{1,r}$
is a proper subset of
$\mathbb {Q} \mathbb {A}_r$
. However, Lemma 1.5 shows that these two classes are actually equivalent. Thus, in light of Theorem 1.3, we have an analogous model for
$\mathbb Q \mathbb A_r$
in Theorem 1.6 whose proof is given below.
Proof of Theorem 1.6.
The converse is straightforward. We assume that
$T \in \mathbb {Q} \mathbb {A}_r$
. We have by Lemma 1.5 that
$rT \in C_{1, r^2}$
. It follows from Theorem 1.3 that there is a Hilbert space
$\mathcal {K} \supseteq {\mathcal {H}}$
, an
$A_{r^2}$
-unitary
$\tilde {N}$
on
$\mathcal K$
and a self adjoint unitary
$F $
on
$\mathcal K$
such that
$$ \begin{align} f(rT)=P_{\mathcal{H}} \bigg(f_0(\tilde{N})f_1(\tilde{N})^{-1}f_2(F\tilde{N}F)^{-1}\bigg) \bigg|_{\mathcal{H}} \end{align} $$
for every rational function
$f=f_0/ f_1f_2$
with zeros of
$f_1$
and
$f_2$
lying in
$\mathbb {C} \setminus \overline {\mathbb {D}}$
and
$r^2\mathbb {D}$
respectively. Let
$g=p/ q_1q_2$
with zeros of
$q_1$
and
$q_2$
in
$\mathbb {C} \setminus r^{-1} \overline {\mathbb {D}}$
and
$r\mathbb {D}$
respectively. We define
$$\begin{align*}f(z)=g(r^{-1}z)=\frac{p(r^{-1}z)}{q_1(r^{-1}z)q_2(r^{-1}z)}, \end{align*}$$
which is holomorphic on
$\overline {A}_{r^2}$
with zeros of
$q_1(r^{-1}z)$
and
$q_2(r^{-1}z)$
lying in
$\mathbb {C} \setminus \overline {\mathbb {D}}$
and
$r^2\mathbb {D}$
respectively. We have by (4.1) that
$$\begin{align*}g(T)=f(rT)=P_{\mathcal{H}} \bigg(p(r^{-1}\tilde{N})q_1(r^{-1}\tilde{N})^{-1}q_2(r^{-1}F\tilde{N}F)^{-1}\bigg) \bigg|_{\mathcal{H}}. \end{align*}$$
Let
$N=r^{-1} \tilde {N}$
. Then N is a normal operator and we have that
$$\begin{align*}\sigma(N)=\{r^{-1}\lambda : \lambda \in \sigma(\tilde{N})\}\subseteq \{r^{-1}\lambda : |\lambda|=1 \ \text{or} \ |\lambda|=r^2\}=\partial \overline{\mathbb{A}}_r. \end{align*}$$
The proof is now complete.
Since we have an equivalence of the two classes
$C_{1,r}$
and
$\mathbb {Q} \mathbb {A}_r$
by Lemma 1.5, we have a model in Theorem 1.8 for
$C_{1,r}$
analogous to the model for
$\mathbb {Q} \mathbb {A}_r$
due to McCullough and Pascoe. We present a brief proof to this below.
Proof of Theorem 1.8.
The converse is trivial. Let us assume that
$T \in C_{1, r}$
. By Lemma 1.5,
$r^{-1/ 2} T \in \mathbb {Q} \mathbb {A}_{\sqrt {r}}$
. It follows from Theorem 1.2 that there is a unitary U on some larger Hilbert space
$\mathcal {K} \supseteq {\mathcal {H}}$
, an invertible operator
$\tilde {J}$
on
$\mathcal {K}$
such that
$(r^{-1/ 2}T)^n=P_{\mathcal {H}}\tilde {J}^n|_{\mathcal {H}}$
for all
$n \in \mathbb {Z}$
and that
$$\begin{align*}\tilde{J}=U\begin{bmatrix} r^{1/ 2}I_{\mathcal{K}_{0}} & 0\\ 0 & r^{-1/ 2}I_{\mathcal{K}_{1}} \end{bmatrix}, \end{align*}$$
where U is a unitary and
$\mathcal {K}=\mathcal {K}_0 \oplus \mathcal {K}_1$
. Take
$J=r^{1/ 2}\tilde {J}$
and the desired conclusion follows.
Theorem 1.8 gives a model for an operator in the
$C_{1,r}$
class. Therefore, the model operator J as in Theorem 1.8 cannot be an
$A_r$
-unitary. The reason is obvious; an
$A_r$
-unitary can provide a model for the
$A_r$
-contractions and the
$C_{1,r}$
class is strictly bigger than that of the
$A_r$
-contractions. So, a natural question arises: when the model operator J becomes an
$A_r$
-unitary so that the initial operator
$T \in C_{1,r}$
becomes an
$A_r$
-contraction ? Below we provide a necessary and sufficient condition for the same.
Proposition 4.1 Let J acting on a Hilbert space
$\mathcal {K}$
be as in Theorem 1.8. Then the following are equivalent:
-
(1) J is an
$A_r$
-unitary ; -
(2) J is normal ;
-
(3)
$UJ_r=J_rU$
where
$ J_r=\begin {bmatrix} rI_{\mathcal {K}_0} & 0 \\ 0 & I_{\mathcal {K}_1} \end {bmatrix}$
.
Proof
$(1) \implies (2)$
follows trivially. We shall prove
$(2) \implies (3) \implies (1)$
.
$(2) \implies (3)$
. Let
$J=UJ_r$
be normal. Then
$J^*J=JJ^*$
implies that
$UJ_r^2=J_r^2U$
. Since
$J_r$
is self-adjoint, one can prove that
$UJ_r=J_rU$
by an application of spectral theorem as discussed in the proof of Theorem 1.4.
$(3) \implies (1)$
. Let
$UJ_r=J_rU$
and let
$U=\begin {bmatrix} U_{11} & U_{12} \\ U_{21} & U_{22} \end {bmatrix}$
with respect to
$\mathcal {K}=\mathcal {K}_0 \oplus \mathcal {K}_1$
. Then
$$\begin{align*}0=UJ_r-J_rU=\begin{bmatrix} rU_{11} & U_{12} \\ rU_{21} & U_{22} \end{bmatrix}-\begin{bmatrix} rU_{11} & rU_{12} \\ U_{21} & U_{22} \end{bmatrix}=\begin{bmatrix} 0 & (1-r)U_{12} \\ (r-1)U_{21} & 0 \end{bmatrix}\,, \end{align*}$$
which is possible if and only if
$U_{12}=U_{21}=0$
as
$r<1$
. Hence,
$U_{11}$
and
$U_{22}$
are unitaries on
$\mathcal {K}_0$
and
$\mathcal {K}_1$
respectively. Furthermore, we have that
$$\begin{align*}J=\begin{bmatrix} rU_{11} & 0 \\ 0 & U_{22} \end{bmatrix}. \end{align*}$$
It follows from Theorem 3.1 that J is an
$A_r$
-unitary which completes the proof.
Concluding remark
The model operators by Bello-Yakubovich [Reference Bello and Yakubovich4] for
$C_{\alpha }$
class or by McCullough-Pascoe [Reference McCullough and Pascoe12] for
$\mathbb Q \mathbb A_r$
cannot be normal or subnormal operators as a subnormal operator in
$C_{1,r}$
is an
$A_r$
-contraction by Proposition 2.10 and a subnormal operator in
$\mathbb Q \mathbb A_r$
has
$\overline {\mathbb A}_r$
as a spectral set by Proposition 2.11. The chain
$\mathcal A_r \subsetneq C_{\alpha } \subsetneq C_{1,r}$
clearly shows that an operator in
$\mathcal A_r$
cannot be a model for
$C_{1,r}$
. Also Example 2.12 confirms that the operators having
$\overline {\mathbb A}_r$
as a spectral set is properly contained in
$\mathbb Q \mathbb A_r$
and thus cannot be a model for the
$\mathbb Q \mathbb A_r$
. However, our models for both
$C_{1,r}$
and
$\mathbb Q \mathbb A_r$
consist of a pair of normal operators having their spectrums on the boundary of the annuli
$A_r$
and
$\mathbb A_r$
respectively.
Data availability statement
(1) Data sharing is not applicable to this article as no datasets were generated or analysed during the current study.
(2) In case any datasets are generated during and/or analysed during the current study, they must be available from the corresponding author on reasonable request.
Competing interests
There are no competing interests.
